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# 数学代写|微积分代写Calculus代考|MATH1051 Estimating sums using the integral test

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## 数学代写|微积分代写Calculus代考|Estimating sums using the integral test

In example 3 we reasoned that because $\int_1^{\infty} \frac{1}{x^3} d x$ converges, $\sum_{n=1}^{\infty} \frac{1}{n^3}$ also converges. However, we did not compute the sum of the series. This is because the value of the integral and the sum of the series are not the same. In fact, although
$$\int_1^{\infty} \frac{1}{x^3} d x=-\left.\frac{1}{2 x^2}\right|1 ^{\Omega}=-\frac{1}{2 \Omega^2}-\left(-\frac{1}{2}\right) \approx \frac{1}{2},$$ the sum of the series is obviously larger: $$\sum{n=1}^{\infty} \frac{1}{n^3}=1+\frac{1}{8}+\frac{1}{27}+\cdots>1 .$$
This relationship is consistent with figure 1 , from which we concluded
$$\sum_{n=1}^{\infty} a_n>\int_1^{\infty} f(x) d x .$$
However, figure 2 gives an upper bound for the sum of the series. Although we did not include $a_1$ when we noticed that
$$\sum_{n=2}^{\infty} a_n<\int_1^{\infty} f(x) d x,$$
including $a_1$ results in
$$\sum_{n=1}^{\infty} a_n<a_1+\int_1^{\infty} f(x) d x .$$

## 数学代写|微积分代写Calculus代考|Proof of the sum of powers approximation formula

Consider $\sum_{n=1}^{\infty} n^2=1+4+9+16+\cdots$. Although $f(x)=x^2$ has the properties that $f(n)=n^2$ for every natural number $n, f$ is positive on $[1, \infty)$, and $f$ is continuous on $[1, \infty)$, the integral test still does not apply because $f$ is increasing on $[1, \infty)$, not decreasing. We can try the test for divergence (which is the first thing we should try anyway, even before the integral test). In doing so, we have
$$\lim {n \rightarrow \infty} n^2=\Omega^2 \doteq \infty \neq 0,$$ and the series diverges. If this series looks vaguely familiar, there is a reason for it. We used the capital omegath partial sum of this series, $\sum{n=1}^{\Omega} n^2$, when evaluating omega sums (although we used the variable $k$ instead of $n$ ). As you may recall, a key step in evaluating omega sums is the use of the sum of powers approximation formula that, when applied in this case, is $\sum_{k=1}^{\Omega} k^2 \approx \frac{\Omega^3}{3}$

The proof of the sum of powers approximation formula is similar to the proof of the integral test. Figure 3 is similar to figures 1 and 2 , but with an increasing function instead of a decreasing function.

## 数学代写|微积分代写Calculus代考|Estimating sums using the integral test

$$\int_1^{\infty} \frac{1}{x^3} d x=-\frac{1}{2 x^2} \mid 1^{\Omega}=-\frac{1}{2 \Omega^2}-\left(-\frac{1}{2}\right) \approx \frac{1}{2},$$

$$\sum n=1^{\infty} \frac{1}{n^3}=1+\frac{1}{8}+\frac{1}{27}+\cdots>1 .$$

$$\sum_{n=1}^{\infty} a_n>\int_1^{\infty} f(x) d x .$$

$$\sum_{n=2}^{\infty} a_n<\int_1^{\infty} f(x) d x$$

$$\sum_{n=1}^{\infty} a_n<a_1+\int_1^{\infty} f(x) d x .$$
formula

## 数学代写|微积分代写Calculus代考|Proof of the sum of powers approximation formula

$$\lim n \rightarrow \infty n^2=\Omega^2 \doteq \infty \neq 0,$$

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